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(D) Let the origin be at the circumcentre $P$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}|

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$\overrightarrow{PQ}$

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(D) Let the origin be at the circumcentre $P$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$,where $R$ is the circumradius.The position vector of the orthocentre $Q$ is given by $\vec{q} = \vec{a} + \vec{b} + \vec{c}$.Since $P$ is the origin,the position vector of $P$ is $\vec{p} = \vec{0}$.We need to find $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$.This is equal to $(\vec{a} - \vec{p}) + (\vec{b} - \vec{p}) + (\vec{c} - \vec{p}) = \vec{a} + \vec{b} + \vec{c} - 3\vec{p}$.Since $\vec{p} = \vec{0}$,this simplifies to $\vec{a} + \vec{b} + \vec{c}$.Since $\vec{q} = \vec{a} + \vec{b} + \vec{c}$ and $\vec{p} = \vec{0}$,we have $\vec{q} - \vec{p} = \vec{a} + \vec{b} + \vec{c}$.Thus,$\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{PQ}$.

If the points $P$ and $Q$ are respectively the circumcentre and the orthocentre of a $\triangle ABC$,then $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}$ is equal to

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